Question: Find $\lim_{x\to\infty} \left(1-\dfrac{4}{x}\right)^{ x}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $\dfrac14$ (Choice C) C $\dfrac{1}{e^4}$ (Choice D) D The limit doesn't exist.
Answer: Taking $x$ to $\infty$ in $\left(1-\dfrac{4}{x}\right)^{x}$ results in the indeterminate form $1^{^{\infty}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=\left(1-\dfrac{4}{x}\right)^{x}$, we will find $\lim_{x\to \infty}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to \infty}y$. $\ln(y) =\dfrac{\ln\left(1-\dfrac{4}{x}\right)}{x^{-1}}$ Taking $x$ to $\infty$ in $\dfrac{\ln\left(1-\dfrac{4}{x}\right)}{x^{-1}}$ results in the indeterminate form $\dfrac{0}{0}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to \infty}\ln(y) \\\\ &=\lim_{x\to \infty}\dfrac{\ln\left(1-\dfrac{4}{x}\right)}{x^{-1}} \\\\ &=\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}\left[\ln\left(1-\dfrac{4}{x}\right)\right]}{\dfrac{d}{dx}[x^{-1}]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to \infty}\dfrac{\left(1-\dfrac{4}{x}\right)^{-1}\cdot\left(\dfrac4{x^2}\right)}{-\dfrac{1}{x^2}} \\\\ &=\lim_{x\to \infty}\dfrac{4{x^2}}{\left(1-\dfrac{4}{x}\right)\cdot-{x^2}} \\\\ &=-4 \gray{\text{The leading terms are the coefficients of $x^2$}} \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}\left[\ln\left(1-\dfrac{4}{x}\right)\right]}{\dfrac{d}{dx}[x^{-1}]}$ actually exists. We found that $\lim_{x\to \infty}\ln(y)=-4$, which means $\lim_{x\to \infty}y=\dfrac{1}{e^4}$. [Why?] In conclusion, $\lim_{x\to\infty} \left(1-\dfrac{4}{x}\right)^{ x}=\dfrac{1}{e^4}$.